Your question is very general, and so is this answer.

When a power plant creates power like the Hoover Dam, it can provide 2.07 GW of electrical power. My question is what does this mean? I assume from Faraday’s law that the induced voltage across the generator coil produces an current, and this combination (P = VI) is the actual power but I'm sure that my thinking is naïve. Can someone roughly sketch out how the electrical power of a power plant is computed?

## From a mechanical perspective

"2.07 GW" means that the peak output of the power plant is 2.07 GW. This is most likely a series of smaller units, say 20 × 100 MW units = 2.0 GW.

The generator is a converter of mechanical energy into electrical energy. So to generate 2.07 GW of electrical energy, an equivalent amount of mechanical energy has to be provided. In the case of the Hoover Dam, the mechanical energy is provided by water falling to a lower elevation, giving up its gravitational potential energy in the process.

From this perspective, you can think of the maximum electrical output power of a power plant as the maximum rate at which it can convert mechanical energy into electrical energy, factoring in the efficiency of the conversion process.

The rate at which mechanical power is generated is a mechanical engineer's problem. For a hydroelectric station, the mechanical power would depend on the water pressure, turbine size, and various design parameters. For a wind turbine, the mechanical power would be set by the radius of the blades. And so on.

## From an electrical perspective

Yes, the electrical energy produced follows Faraday's Law and Ohm's Law, though for an AC system, V and I are sinusoids which may not be in phase, and P ≠ VI. Rather, apparent power (volt-amperes) S = VI, and real power (watts) P = VI cos ɸ.

Other complications include electrical losses (per Ohm's law) and magnetic losses (eddy currents induced in metallic parts).

If possible, what kind of voltages and currents are power plants producing before the Step Up transformers? Of course, this varies from one power plant to another.

## Regarding typical voltages

In my experience, small generators (i.e. diesel gen-sets) generate directly at the utilisation voltage, say 415 V here in Australia.

Larger power station units generate at a medium voltage like 11kV before stepping up to transmission voltage, i.e. 132 kV.

I imagine a medium voltage like 11kV is preferred vs. a high voltage like 33kV, because less insulation is required on the windings and the rotating parts may be physically lighter.

## Regarding typical currents

An aeroderivative gas turbine, i.e. the General Electric LM6000, is typically rated about 45 MW and might have a 60 MVA alternator attached to it. Calculation of the three-phase line current at 11kV is left as an exercise to the reader. Don't forget your √3.

A coal power station unit might be rated 400 MVA at 22kV. See "Tarong Power Station" in QLD, Australia, which consists of four large units like this. Again, calculation of the line current is left as an exercise.

Note: I am at home and hence don't have access to my reference material at work. The above numbers are indicative, so treat them with a grain of salt.

If you are curious as to the exact operating principles and theory of an AC generator, I would encourage you to look up a textbook on electric machinery. My personal favourite is Mulukutla Sarma's *Electric Machines*. Check your university library for a copy.

KVL can be more generally stated as: the electric field is a conservative vector field. This can be expressed by the path integral:
\begin{align}
\int_{p1}^{p2} \vec{E} \cdot d\vec{l} = V_{p2} - V_{p1}
\end{align}
I.E. the integral is "path independent", and the result is the same no matter what path you take (so long as you start at p1 and end at p2). This can be 0, but it doesn't have to be. If \$p1=p2\$ (taking a closed path),
\begin{align}
\oint \vec{E} \cdot d\vec{l} = 0
\end{align}

This is why KVL even applies: KVL states that in any **closed** loop, the voltage across it is 0. It doesn't mean that the voltage must be 0 between any two points on that loop. It doesn't matter if there is any conductive path between two points. Having an ideal wire is just an *extra* statement that two distinct points are at the same voltage.

## Best Answer

Almost all are piezo-electric. Check this Wikipedia page for info:

Piezo ignition is a type of ignition that is used in portable camping stoves, gas grills and some lighters. It consists of a small, spring-loaded hammer which, when a button is pressed, hits a crystal of PZT or quartz crystal. Quartz is piezoelectric, which means that it creates a voltage when deformed. This sudden forceful deformation produces a high voltage and subsequent electrical discharge, which ignites the gas.

And as Christopher Biggs commented: